The equation of the vertical parabola in vertex form is written as
Where (h, k) are the coordinates of the vertex and p is the focal distance.
The directrix of a parabola is a line which every point of the parabola is equally distant to this line and the focus of the parabola. The vertex is located between the focus and the directrix, therefore, the distance between the y-coordinate of the vertex and the directrix represents the focal distance.
Using this value for p and (3, 1) as the vertex, we have our equation
Answer:
y = 3sin2t/2 - 3cos2t/4t + C/t
Step-by-step explanation:
The differential equation y' + 1/t y = 3 cos(2t) is a first order differential equation in the form y'+p(t)y = q(t) with integrating factor I = e^∫p(t)dt
Comparing the standard form with the given differential equation.
p(t) = 1/t and q(t) = 3cos(2t)
I = e^∫1/tdt
I = e^ln(t)
I = t
The general solution for first a first order DE is expressed as;
y×I = ∫q(t)Idt + C where I is the integrating factor and C is the constant of integration.
yt = ∫t(3cos2t)dt
yt = 3∫t(cos2t)dt ...... 1
Integrating ∫t(cos2t)dt using integration by part.
Let u = t, dv = cos2tdt
du/dt = 1; du = dt
v = ∫(cos2t)dt
v = sin2t/2
∫t(cos2t)dt = t(sin2t/2) + ∫(sin2t)/2dt
= tsin2t/2 - cos2t/4 ..... 2
Substituting equation 2 into 1
yt = 3(tsin2t/2 - cos2t/4) + C
Divide through by t
y = 3sin2t/2 - 3cos2t/4t + C/t
Hence the general solution to the ODE is y = 3sin2t/2 - 3cos2t/4t + C/t
Answer:
-2/3
Step-by-step explanation:
Where 2 lines are perpendicular, the slope of one = - 1 / slope of the other.
Thus the slope of the red line is - 1 / 3/2
= -2/3
Adult tickets = A
Children tickets = C
A×4. C×2
A×3. C×4
A×2. C×6
A×1. C×8
